(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Dear Friends,

Given the differential equation

[tex]\frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)[/tex]

with the condition that x(0) = 0

Then find the largest possible solution (this is how its stated) if either

[tex]\alpha_{1} = 1/2[/tex]

[tex]\alpha_{2} = 1[/tex]

3. The attempt at a solution

Don't I treat the above as a seperable differential equation?

Such that

[tex]{(\frac{1}{1+x^2})}^{-1})} \frac{dx}{dt} = {({\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)})}^{-1}[/tex]

[tex]1+x^2 dx = \frac{2}{\alpha \cdot \pi \cdot cos(t)} dt[/tex]

[tex]\int x^2 dx = \int \frac{2}{\alpha \cdot \pi \cdot cos(t)} -1 \ dt + C[/tex]

[tex] \frac{x^3}{3} = \frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi}[/tex]

[tex] x^3 = 3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})[/tex]

[tex] x(t) = (3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})^{\frac{1}{3}}[/tex]

if I then insert x(0) = 0.

I get C = 0. Then this means that max solution is 0??

I just want to be sure. Can anybody please look at my result at then inform me. Have done this correctly?

Thank You in Advance

Fred

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# Homework Help: Differential equation problem

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